All categories

3 years ago

Briefly explain earth's daily rotation and annual orbit, defining the terms ecliptic plane and axis tilt.

1 answer:

8 0

<span>The movement around the axis of the earth occurs once every 24 hours and it does so from a start point to finish in a term we can call complete revolution and it does this by day by day turn, every 24 hours. The yearly circle where the earth moves round the sun , however is a 365 transformation around the sun or ecliptic plane. It does this with the earth tilted at 23.5 degrees, off the opposite line to the ecliptic plane.</span>

You might be interested in

A 30.0 g mass of iron at 24.5°C is heated to 45.0°C. The theoretical specific

diamong [38]

Answer:

276.135 J

Explanation:

Given that:

mass of Fe = 30.0 g

initial temperature = 24.5°C

final temperature = 45.0°C

specific heat of Fe = 0.449 J/g°C

We can determine the thermal energy added by using the formula;

Q = mcΔT

Q = 30.0g × 0.449 J/g°C × (45.0 - 24.5)°C

Q = 276.135 J

8 0

3 years ago

A small 12.00 g plastic ball is suspended by a string in a uniform, horizontal electric field. If the ball is in equilibrium whe

notsponge [240]

Answer:

$Q = \frac{0.068}{E}$

where E = electric field intensity

Explanation:

As we know that plastic ball is suspended by a string which makes 30 degree angle with the vertical

So here force due to electrostatic force on the charged ball is in horizontal direction along the direction of electric field

while weight of the ball is vertically downwards

so here we have

$QE = F_x$

$mg = F_y$

since string makes 30 degree angle with the vertical so we will have

$tan\theta = \frac{F_x}{F_y}$

$tan30 = \frac{QE}{mg}$

$Q = \frac{mg}{E}tan30$

$Q = \frac{0.012\times 9.81}{E} tan30$

$Q = \frac{0.068}{E}$

where E = electric field intensity

5 0

3 years ago

Which atomic model proposed that electrons move in specific orbits around the nucleus of an atom

BabaBlast [244]

Bohr's atomic model

8 0

3 years ago

Un ladrillo se le imparte una velocidad inicial de 6m/s en su trayectoria hacia abajo. ¿cual sera su velocidad final despues de

marshall27 [118]

Saludos!

Respuesta:

28,64 m/s.

Explicación:

Datos:

Altura o distancia recorrida: 40 m
Vo: 6 m/s
Aceleración de la gravedad: 9,81 m/s²

El ejercicio puede ser resuelto facilmente utilizando la siguiente formula, sin embargo es posible realizarlo utilizando formulas diferentes.

Entonces tenemos que:

$Vf ^{2} -Vo ^{2} =2 x g x h$

Es importante saber que al estar lanzando el ladrillo hacia abajo, el sentido del movimiento sigue el sentido de la gravedad, es decir es necesario que tomes el valor de la gravedad como positivo (+) y no negativo (-) como normalmente se usa.

Sustituyendo tenemos que:

$Vf x^{2} =(6 m/s) ^{2} +(2)x(9,81 m/s ^{2})x(40m) \\ Vf ^{2} =820,8 m^{2} /s ^{2} \\ \sqrt{Vf ^{2} } = \sqrt{820,8m^{2}/s ^{2} } \\ Vf=28,64 m/s$

Que tengas un buen día!

6 0

3 years ago

Geocentric theory:

grandymaker [24]

I believe d all of the above

3 0

3 years ago